Suppose we have two related questions:
(a) A family has 2 children. What is the probability of having 2 girls for the family?
(b) A family has 2 children. What is the probability of having 2 girls given that one of the children is girl?
Question (b) has a conditional probability element, while (a) doesn't. To answer (a), we need to list 4 possible cases of having 2 kids: gg, bg, gb, and bb. (b = boy, g = girl.) The probability is thus 1/4 for (a).
We only have 3 possible cases for (b): gg, bg, gb, so that the probability is 1/3.
The notion of conditional probability is the first hurdle for students taking the probability and statistics course I teach this term.
We write P(B | A) for the probability of B to occur, given that A occurs,
We write P(B | A) for the probability of B to occur, given that A occurs,
P(B | A) = P(B ∩ A)/P(A),
where B ∩ A means both events A and B occur.
If we want to apply this formula to answer part (b), then P(B ∩ A) = 1/4 since there is only one case (gg); P(A) = 3/4 since having 1 girl can happen with first child (gb), second child (bg), or both girls (gg).
The formula P(B | A) = P(B ∩ A)/P(A) is counterintuitive though, so I would write it as
P(B ∩ A) = P(B | A) P(A).
That is, the probability of having A and B to occur is equal to (i) having A to occur, and (ii) having B to occur given A occurs. The last expression gives more clarity since if A and B are statistically independent, then
P(B | A) = P(B),
so that when A and B are independent, P (B ∩ A) = P(B) P(A).
Depending on available data, any P(B ∩ A), P(B | A), or P(A) can be computed, but clearly we need to know two in order to solve for the third.
Another useful probability formula is
P(A ∩ B) = P(A) + P(B) – P(A ∪ B),
where A ∪ B means either event A or B occurs.
The conditional probability can be used to answer an unconditional probability. For example, the probability of picking for the first time an odd number from the 5 numbers: 1, 2, 3, 4, 5 is 3/5 since there are 3 odd numbers out of 5. What is now the unconditional probability for picking an odd number for the second time (thus, regardless of the outcome of the first pick)?
The conditional probability can be used to answer an unconditional probability. For example, the probability of picking for the first time an odd number from the 5 numbers: 1, 2, 3, 4, 5 is 3/5 since there are 3 odd numbers out of 5. What is now the unconditional probability for picking an odd number for the second time (thus, regardless of the outcome of the first pick)?
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